Question

Given an aritmetic progression 2,6,10,14.. find the smallest value of n such that the sum of the first n terms is greater than 200.​

Answer 1

Hello,

Explanation:

Using n terms, the arithmetic sequence will be:

2,6,10,14,...,?

=0*4+2,1*4+2,2*4+2,3*4+2,...,(n-1)*4+2

s=(0*4+2)+(1*4+2)+(2*4+2)+(3*4+2)+...+(n-1)*4+2 >200

(2+2+2+...+2)+4(0+1+2+...+(n-1)) > 200

2*n+4*(0+(n-1))*n/2 >200

2n+2(n²-n) > 200

2n² >200

n² > 100

n >10

the smallest value of n is 11.

11 th term is (11-1)*4+2=42

sequence={2,6,10,14,...,42} which sum is 2*11²=242

Answer 2

Without thinking too hard, by adding the progression of number until we reach >200 and counting what n value we have we can solve this question.

The rule of this progression is to + 4 every time.

2 + 6 + 10 + 14 + 18 + 22 + 26 + 30 + 34 + 38 + 42

This will give us 242, which is greater than 200. There are 11 numbers here, so the n value is 11.